07-Str-A1 · December 2013

Question 3 of 8: Vertical deflection of a non-prismatic beam by virtual work

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Str-A1 Elementary Structural Analysis. Three hours, CLOSED BOOK, an approved Sharp or Casio calculator permitted. Six questions constitute a complete paper: candidates answer ALL of Questions 1 to 5 and ONE of Questions 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are solved below, because the three alternatives test different topics and the whole set is useful for study.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3 to 5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8 to 9 (deflections and virtual work), Ch. 11 to 12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (deflections by virtual work), Ch. 8 (influence lines), Ch. 16 to 17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text after which this exam code is named.
Canadian design context: CSA S16 Design of Steel Structures, CSA A23.3 Design of Concrete Structures and the National Building Code of Canada. This is an analysis paper, so no code clause is needed to answer it, but every result below is expressed in the SI units those documents use.

Sign conventions used throughout. For beams and for each individual frame member, shear is positive when the resultant of the forces to the left of (or below) a section acts upward, and bending moment is positive when it sags the member, that is when it puts the underside of a beam or the inside face of a frame member in tension. Member end moments in the slope-deflection work of Question 5 are clockwise-positive on the member, which is the usual convention for that method. Truss forces are quoted as tension or compression rather than by sign. Reactions are drawn in blue and applied loads in red on every figure.

A note on reading this paper. The figures are hand drawn and carry no dimensions that are not on the drawings, so the geometry of every structure below was read from the support symbols one at a time. Two readings materially affect the answers and are flagged where they arise: the plane on which the roller of Question 2(c) bears, and the position of the 50 kN load in Question 8.

Question 3: Vertical deflection of a non-prismatic beam by virtual work (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Span A to C (pin at A, roller at C)	12 m, made of AB = 6 m and BC = 6 m
Overhang C to D	3 m, free end at D
Flexural rigidity, AB	$EI = 8.0 \times 10^{3}$ kN·m$^{2}$
Flexural rigidity, BC and CD	$3EI = 2.4 \times 10^{4}$ kN·m$^{2}$
Point load at B	12 kN downward
Point load at D	8 kN downward

Find. The vertical deflection of point B, with its direction.

The non-prismatic beam: AB has rigidity EI while BC and the CD overhang have 3EI.

Real bending moment diagram (left) and the virtual moment diagram (right) for a 1 kN downward load at B.

Approach. Use the unit-load form of virtual work. Draw the real bending moment diagram, then the moment diagram for a 1 kN virtual load placed at B in the direction sought, and integrate their product divided by the local flexural rigidity over each of the three segments.

Find the real reactions. Measure $x$ from A. Taking moments about A with the roller at $x = 12$ m, $$R_{C}\,(12) = 12\,(6) + 8\,(15) = 72 + 120 = 192 \;\Rightarrow\; R_{C} = 16\ \text{kN}\uparrow$$ $$R_{A} = 12 + 8 - 16 = 4\ \text{kN}\uparrow$$
Write the real moment in each segment. Working from the left, $$\begin{aligned} M &= 4x \quad (0 \le x \le 6) \\ M &= 72 - 8x \quad (6 \le x \le 12) \\ M &= 8x - 120 \quad (12 \le x \le 15) \end{aligned}$$ so $M_{B} = +24$ kN·m, $M_{C} = -24$ kN·m and $M_{D} = 0$ as the free end requires.
Apply the virtual unit load. Put 1 kN downward at B on the same beam. Its reactions are $r_{A} = 0.5$ kN and $r_{C} = 0.5$ kN, so $$\begin{aligned} m &= 0.5x \quad (0 \le x \le 6) \\ m &= 6 - 0.5x \quad (6 \le x \le 12) \\ m &= 0 \quad (12 \le x \le 15) \end{aligned}$$ The overhang contributes nothing, because a load applied between the supports produces no moment in a free-ended overhang. That is a useful check on the virtual diagram before any integration is attempted.
Integrate segment AB, where the rigidity is EI. $$\int_{0}^{6}\frac{Mm}{EI}\,\mathrm{d}x = \frac{1}{EI}\int_{0}^{6}(4x)(0.5x)\,\mathrm{d}x = \frac{2}{EI}\Big[\frac{x^{3}}{3}\Big]_{0}^{6} = \frac{144}{EI}$$ Both diagrams are triangles with their apex at B, so the same number follows from the product-integral table as $\tfrac{1}{3}(24)(3)(6)/EI = 144/EI$.
Integrate segment BC, where the rigidity is 3EI. Here the real moment falls from $+24$ to $-24$ kN·m while the virtual moment falls from $+3$ to zero, so the product changes sign inside the segment and the integral must be taken algebraically: $$\int_{6}^{12}\frac{Mm}{3EI}\,\mathrm{d}x = \frac{1}{3EI}\int_{6}^{12}(72 - 8x)(6 - 0.5x)\,\mathrm{d}x = \frac{1}{3EI}\Big[\frac{4x^{3}}{3} - 42x^{2} + 432x\Big]_{6}^{12} = \frac{72}{3EI} = \frac{24}{EI}$$
Add the segments and substitute EI. Segment CD contributes nothing because $m = 0$ there, so $$\delta_{B} = \frac{144 + 24}{EI} = \frac{168}{8.0 \times 10^{3}} = \boxed{0.0210\ \text{m} = 21.0\ \text{mm downward}}$$ The sign is positive, which means the deflection is in the direction of the applied unit load, that is downward. Had the stiffer 3EI extended over the whole span the answer would have been $168/3EI$ or 7.0 mm, so the soft segment AB accounts for nearly all of the movement.

Quantity	Value
Reaction at A	4.0 kN upward
Reaction at C	16.0 kN upward
Real moment at B	+24.0 kN·m
Real moment at C	−24.0 kN·m
Virtual moment at B for 1 kN at B	+3.00 m
Contribution of AB	$144/EI$
Contribution of BC	$24/EI$
Vertical deflection at B	21.0 mm downward

Topic. Deflection of a non-prismatic beam by the unit-load (virtual work) method.

Key relations. $1 \cdot \delta = \displaystyle\int \frac{Mm}{EI}\,\mathrm{d}x$, evaluated segment by segment with the local $EI$.

Why this works. Virtual work turns a deflection into an integral of two bending moment diagrams, one from the real loads and one from a dummy unit load placed at the point and in the direction whose movement is wanted. Because the rigidity appears inside the integral, a step change in section is handled simply by splitting the integral at the step; no special theory is needed. The method also tells you where the flexibility lives: here the 6 m of plain EI next to the pin supplies 144 of the 168 units, six times as much as the equally long but three times stiffer segment beyond it.

Common pitfall. Dividing by the wrong rigidity after the section changes, and integrating the product of two diagrams whose sign differs over part of the segment as if it were a single positive area. In BC the real moment crosses zero at $x = 9$ m; using an area formula that assumes one sign gives about $60/EI$ instead of $24/EI$.

Source. R. C. Hibbeler, Structural Analysis, 10th ed., Sect. 8.4 and Ch. 9 (virtual work for beams); A. Kassimali, Structural Analysis, 6th ed., Sect. 7.4–7.6.

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